In Young's double slit experiment, two slits 0.12 mm apart are illuminated by monochromatic light of wavelength 480 nm. The screen is 1.0 m away from the slits. The distance of the second bright fringe from the central maximum will be

$8 × 10^{-3} m$

The correct answer is Option (4) → $8 × 10^{-3} m$

Given:

Slit separation, $d = 0.12 \ \text{mm} = 1.2 \times 10^{-4} \ \text{m}$

Wavelength, $\lambda = 480 \ \text{nm} = 4.8 \times 10^{-7} \ \text{m}$

Distance to screen, $L = 1.0 \ \text{m}$

Order of bright fringe, $m = 2$ (second bright fringe)

Fringe separation: $\Delta y = \frac{\lambda L}{d}$

$\Delta y = \frac{4.8 \times 10^{-7} \cdot 1}{1.2 \times 10^{-4}} = 4 \times 10^{-3} \ \text{m} = 4 \ \text{mm}$

Distance of second bright fringe from central maximum: $y = m \Delta y = 2 \cdot 4 \ \text{mm} = 8 \ \text{mm}$

Distance from central maximum: $y = 8 \ \text{mm}$